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Determine the pH of 8.50×10−2 M KCN?

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  • 2 months ago

    CN- acts as a base by the equilibrium:

    CN- + H2O <--> HCN + OH-

    Kb = [HCN][OH-]/[HCN]

    Kb can be calculated from Ka of HCN as:Kb = 1.00X10^-14 / Ka

    Kb = 1.00X10^-14 / 6.2X10^-10 = 1.6X10^-5

    In the solution, let [HCN] = [OH-] = x, and [CN-] = 8.50X10^-2 - x. 

    As a first approximation, assume that x will be small compared to 8.50X10^-2. Then,

    Kb = 1.6X10^-5 = x^2/0.0850

    x = 1.2X10^-3 = [OH-]

    pOH = 2.93

    pH = 14.00 - pOH = 11.07

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